For the ellipse, `9x^(2)+16y^(2)=576`, find the semi-major axis, the semi-minor axis, the eccentricity, the coordinates of the foci, the equations of the directrices, and the length of the latus rectum.

Answer

To solve the problem for the ellipse given by the equation \(9x^2 + 16y^2 = 576\), we will follow these steps:

### Step 1: Rewrite the equation in standard form
We start with the equation:
\[ 9x^2 + 16y^2 = 576 \]
To rewrite it in standard form, we divide the entire equation by 576:
\[ \frac{9x^2}{576} + \frac{16y^2}{576} = 1 \]
This simplifies to:
\[ \frac{x^2}{64} + \frac{y^2}{36} = 1 \]
Here, we can identify \(a^2 = 64\) and \(b^2 = 36\).

### Step 2: Find the semi-major and semi-minor axes
From the values of \(a^2\) and \(b^2\):
\[ a = \sqrt{64} = 8 \quad \text{(semi-major axis)} \]
\[ b = \sqrt{36} = 6 \quad \text{(semi-minor axis)} \]

### Step 3: Calculate the eccentricity
The eccentricity \(e\) of the ellipse is given by the formula:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Substituting the values:
\[ e = \sqrt{1 - \frac{36}{64}} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4} \]

### Step 4: Find the coordinates of the foci
For an ellipse with a horizontal major axis, the foci are located at \((\pm ae, 0)\):
\[ \text{Coordinates of foci} = \left(\pm 8 \cdot \frac{\sqrt{7}}{4}, 0\right) = \left(\pm 2\sqrt{7}, 0\right) \]

### Step 5: Find the equations of the directrices
The equations of the directrices for a horizontal ellipse are given by:
\[ x = \pm \frac{a}{e} \]
Calculating this:
\[ \frac{a}{e} = \frac{8}{\frac{\sqrt{7}}{4}} = \frac{8 \cdot 4}{\sqrt{7}} = \frac{32}{\sqrt{7}} = \frac{32\sqrt{7}}{7} \]
Thus, the equations of the directrices are:
\[ x = \pm \frac{32\sqrt{7}}{7} \]

### Step 6: Calculate the length of the latus rectum
The length of the latus rectum \(L\) is given by:
\[ L = \frac{2b^2}{a} \]
Substituting the values:
\[ L = \frac{2 \cdot 36}{8} = \frac{72}{8} = 9 \]

### Summary of Results
- Semi-major axis \(a = 8\)
- Semi-minor axis \(b = 6\)
- Eccentricity \(e = \frac{\sqrt{7}}{4}\)
- Coordinates of the foci: \((\pm 2\sqrt{7}, 0)\)
- Equations of the directrices: \(x = \pm \frac{32\sqrt{7}}{7}\)
- Length of the latus rectum: \(9\)